Reread again and again and again and again…

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fruitmonster97

2492 posts

fruitmonster97

#1 h Feb 2, 2024, 5:27 PM • 3 Y

Y by Danielzh, rickyjwalland, jocaleby1

Every morning, Aya does a $9$ kilometer walk, and then finishes at the coffee shop. One day, she walks at $s$ kilometers per hour, and the walk takes $4$ hours, including $t$ minutes at the coffee shop. Another morning, she walks at $s+2$ kilometers per hour, and the walk takes $2$ hours and $24$ minutes, including $t$ minutes at the coffee shop. This morning, if she walks at $s+\frac12$ kilometers per hour, how many minutes will the walk take, including the $t$ minutes at the coffee shop?

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think4l

344 posts

think4l

#2 h Feb 2, 2024, 5:28 PM

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make two equations of $s$ and $t$ , then plug into final situation to get total time = $\boxed{204}$ .

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gladIasked

648 posts

gladIasked

#3 h Feb 2, 2024, 5:28 PM

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This one took me an embarrassingly long amount of time (this was one of the last problems I solved lol). The answer was $204$ .

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MC413551

2228 posts

MC413551

#4 h Feb 2, 2024, 5:43 PM

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9/s+t/60=4
9/(s+2)+t/60=2+24/60
solve the equation to get 24 minutes as t and 5/2 as s
plug in s+1/2 to get
3 hours and 24 minutes
so 180+24=204

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bluelinfish

1449 posts

bluelinfish

#5 h Feb 2, 2024, 5:53 PM

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Change all units of time to hours. By the given conditions, $\frac{9}{s} + t = 4$ , $\frac{9}{s+2} + t = \frac{12}{5}$ . Subtracting the two equations and solving for the positive solution of $s$ yields $s = \frac{5}{2}$ . This gives $t = \frac{2}{5}$ . Then $\frac{9}{s+1/2} + t = 3 + \frac{2}{5}$ hours, which corresponds to $\boxed{204}$ minutes.

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shendrew7

796 posts

shendrew7

#6 h Feb 2, 2024, 5:54 PM

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Set up the system of equations
$\begin{align*} \frac 9s + \frac{t}{60} &= 4 \\ \frac{9}{s+2} + \frac{t}{60} &= 2.4 \end{align*}$
to get the solution $(s,t) = (2.5, 24)$ . Our answer is then
$60 \cdot \frac{9}{.5 + 2.5} + 24 = \boxed{204}. \quad \blacksquare$

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MC413551

2228 posts

MC413551

#7 h Feb 2, 2024, 6:20 PM

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bluelinfish wrote:

Change all units of time to hours. By the given conditions, $\frac{9}{s} + t = 4$ , $\frac{9}{s+2} + t = \frac{12}{5}$ . Subtracting the two equations and solving for the positive solution of $s$ yields $s = \frac{5}{2}$ . This gives $t = \frac{2}{5}$ . Then $\frac{9}{s+1/2} + t = 3 + \frac{2}{5}$ hours, which corresponds to $\boxed{204}$ minutes.

Just one thing
+t is incorrect even though it works because you need to convert everything in the equation to hours and t is in minutes

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aopsonline2020888

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aopsonline2020888

#8 h Feb 2, 2024, 8:36 PM

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my friend @NoSignOfTheta got 001 for this problem.
can anyone confirm?

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think4l

344 posts

think4l

#9 h Feb 2, 2024, 9:06 PM • 1 Y

Y by fura3334

aopsonline2020888 wrote:

my friend @NoSignOfTheta got 001 for this problem.
can anyone confirm?

yes, i can confirm Click to reveal hidden text

This post has been edited 1 time. Last edited by think4l, Feb 2, 2024, 9:06 PM

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qwerty123456asdfgzxcvb

1086 posts

qwerty123456asdfgzxcvb

#10 h Feb 2, 2024, 9:11 PM

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think4l wrote:

aopsonline2020888 wrote:

my friend @NoSignOfTheta got 001 for this problem.
can anyone confirm?

yes, i can confirm Click to reveal hidden text

yeah i can also confirm

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bossballs1234

31 posts

bossballs1234

#11 h Feb 2, 2024, 9:13 PM

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gladIasked wrote:

This one took me an embarrassingly long amount of time (this was one of the last problems I solved lol). The answer was $204$ .

same… i came back to this four times and spent 40 minutes on it. biggest facepalm

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plang2008

337 posts

plang2008

#12 h Feb 2, 2024, 9:34 PM

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Let $t$ be the number of hours instead of minutes.
At $r$ km/h, Aya is able to walk $9$ km in $\frac {9}{r}$ hours. Therefore, we have the system
$\begin{align*} \frac {9}{s} &= 4 - t \\ \frac {9}{s+2} &= \frac{12}{5} - t \\ \end{align*}$ Subtracting the second from the first gives $\frac{9}{s} - \frac{9}{s+2} = \frac{8}{5}$ . This rearranges to $45 = 4s(s+2)$ , and solving this quadratic gives $s = \frac{5}{2}$ . Plugging this in gives $t = \frac{2}{5}$ .

Therefore, the answer is $60\left(\frac{9}{3} + \frac{2}{5}\right) = \boxed{204}$ .

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navier3072

115 posts

navier3072

#13 h Feb 3, 2024, 10:59 AM

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$\begin{align*} \frac{9}{s}+\frac{t}{60} &= 4 \\ \frac{9}{s+2}+\frac{t}{60} &= 2.4 \\ \frac{18}{s(s+2)} &= 1.6 \implies 4s^2+8s-45=0 \implies (2s-5)(2s+9)=0 \\ \therefore s=2.5 \implies t &=24 \\ \therefore 60 \left( \frac{9}{s+\frac{1}{2}}+\frac{24}{60} \right) &= \boxed{204} \end{align*}$

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elliotcp

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elliotcp

#14 h Feb 15, 2024, 5:41 PM

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There are 2 equations which will help us to find s and t:

1) 9 / s + t / 60 = 4
2) 9 / (s + 2) + t / 60 = 2.4

So from these equations:

9 / s - 9 / (s + 2) = 1.6
18 / s * (s + 2) = 1.6
4 * s * (s + 2) = 45
4 * (s ^ 2) + 8 * s - 45 = 0
Discriminant = 64 - 4 * 4 * (-45) = 784
s1 = 2.5 s2 = -4.5 (We need only the positive one, so s2 is not important)

From the equations, it is clear that t = 24 minutes. Now we need to find the answer from the final equation:

9 / (s + 0.5) * 60 + t = (9 / 3) * 60 + 24 = 180 + 24 = 204 (I multiplied to 60 for converting hour to minute).

So the answer is 204 minutes.

This post has been edited 1 time. Last edited by elliotcp, Feb 15, 2024, 5:42 PM
Reason: just deleted the image

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Mr.Sharkman

500 posts

Mr.Sharkman

#15 h Apr 2, 2024, 2:31 AM

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Solution

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John_Mgr

70 posts

John_Mgr

#16 h Jun 5, 2024, 1:02 AM

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Note: Convert a quantites to a same unit
$Velocity$ = $\frac{Total Distance Covered}{Total Time Taken}$ .
Let t be the time spend in the coffee shop i.e l in hrs
So, $Total Distance Covered$ = $\left(Total Time Taken\right)$ $\times$ $\left(Velocity\right)$ .
First Case:
$V_1$ = $S km/hr$ , $T_1$ = $4 hr-t$ . so $9$ = $S$ $\left(4-t\right)$
Second Case:
$V_2$ = $\left(S+2\right) km/hr$ , $T_2$ = $2 hr+24 min-t$ = $\frac{12}{5}$ hr-t. So, $9$ = $\left(S+2\right)$ $\times$ $\left(\frac{12}{5}-t\right)$
From First and Second Case $\leadsto$
$S$ $\left(4-t\right)$ = $\left(S+2\right)$ $\times$ $\left(\frac{12}{5}-t\right)$
Solving quadratic we get S= $\frac{5}{2}$ or $-\frac{9}{2}$ and $S$ = $-\frac{9}{2}$ is not possible. So $S$ = $\frac{5}{2}$
Third case:
let $X$ =Time spent excluding coffee shop
$V_3$ = $\left(S+\frac{1}{2}\right)$ $\times$ $X$
$9$ = $\left(\frac{5}{2}+\frac{1}{2}\right)$ $\times$ $X$ and $X$ = $3 hrs$
Total Time(T)= $X+t$ $\Rightarrow$ . $T$ = $3 hrs+\frac{2}{5}$ $hrs$ . So, $T$ = $\frac{17}{5}hrs$ or $204 mins$
$Time$ = $\boxed{204 mins}$

This post has been edited 1 time. Last edited by John_Mgr, Jun 5, 2024, 1:24 AM
Reason: Calculation

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wikjay

237 posts

wikjay

#17 h Jun 5, 2024, 4:04 PM

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fruitmonster97 wrote:

Every morning, Aya does a $9$ kilometer walk, and then finishes at the coffee shop. One day, she walks at $s$ kilometers per hour, and the walk takes $4$ hours, including $t$ minutes at the coffee shop. Another morning, she walks at $s+2$ kilometers per hour, and the walk takes $2$ hours and $24$ minutes, including $t$ minutes at the coffee shop. This morning, if she walks at $s+\frac12$ kilometers per hour, how many minutes will the walk take, including the $t$ minutes at the coffee shop?

I kept messing up my units so it took me 50 mins to solve this